Q.Which of the following functions of time represent

Concept: Frequency and Period — A motion is simple harmonic if it can be written as $A \sin(\omega t + \phi)$ (single frequency). Periodic motion repeats after a finite period $T = 2\pi/\omega$; non-periodic motion never repeats.

(a) $\sin\omega t - \cos\omega t = \sqrt{2}\sin(\omega t - \pi/4)$.

This is a single sine function → simple harmonic. Period $T = 2\pi/\omega$.

(b) $\sin^3\omega t = \frac{3}{4}\sin\omega t - \frac{1}{4}\sin 3\omega t$ (using triple-angle identity).

Contains two frequencies ($\omega$ and $3\omega$) → periodic but not simple harmonic. Fundamental period $T = 2\pi/\omega$.

(c) $3\cos(\pi/4 - 2\omega t) = 3\cos(2\omega t - \pi/4)$.

Single cosine → simple harmonic. Period $T = 2\pi/(2\omega) = \pi/\omega$.

(d) $\cos\omega t + \cos 3\omega t + \cos 5\omega t$ — sum of odd multiples of $\omega$. …

The core idea: What makes motion simple harmonic?

Simple harmonic motion (SHM) is defined by a restoring force proportional to displacement. Mathematically, a function $x(t)$ represents SHM if it can be written in the form

where $A$ is constant amplitude, $\omega$ constant angular frequency, and $\phi$ constant phase. The motion is periodic with period $T = 2\pi/\omega$.

A function can be periodic without being simple harmonic — for example, a sum of sine waves with different frequencies (like a Fourier series) repeats after a common period, but it is not a single sine/cosine. A function is non-periodic if it never repeats exactly.

Let’s examine each case.

(a) $\sin\omega t - \cos\omega t$

Step 1: Combine into a single sine.

Recall the identity: $R\sin(\omega t - \phi) = R\sin\omega t \cos\phi - R\cos\omega t \sin\phi$.

We want $\sin\omega t - \cos\omega t = R\sin(\omega t - \phi)$.

Comparing coefficients:

$R\cos\phi = 1$ and $R\sin\phi = 1$.

So $\tan\phi = 1 \Rightarrow \phi = \pi/4$, and $R = \sqrt{1^2 + 1^2} = \sqrt{2}$.

Thus

Step 2: Classify.

This is exactly of the form $A\sin(\omega t + \phi)$ with constant $A = \sqrt{2}$. So it represents simple harmonic motion.

Step 3: Period.

Angular frequency is $\omega$, so period $T = 2\pi/\omega$.

(b) $\sin^{3}\omega t$

Step 1: Expand using a trigonometric identity.

Use $\sin 3\theta = 3\sin\theta - 4\sin^3\theta$. Rearranging:

Here $\theta = \omega t$, so

Step 2: Classify.

This is a sum of two sine waves with frequencies $\omega$ and $3\omega$. It is periodic (both terms repeat when $\omega t$ increases by $2\pi$), but it is not a single sine or cosine — the waveform is distorted. So it is periodic but not simple harmonic.

Step 3: Period.

The fundamental frequency is $\omega$ (the smallest frequency present), so the period is $T = 2\pi/\omega$.

(Check: $\sin 3\omega t$ repeats after $2\pi/3\omega$, but the whole sum repeats only after $2\pi/\omega$.)

(c) $3\cos\left(\frac{\pi}{4} - 2\omega t\right)$

Step 1: Simplify the argument.

Using $\cos(-\theta) = \cos\theta$,

Step 2: Classify.

This is exactly $A\cos(\omega' t + \phi)$ with $A=3$, $\omega' = 2\omega$, $\phi = -\pi/4$. So it is simple harmonic motion.

Step 3: Period.

Angular frequency is $2\omega$, so period $T = 2\pi/(2\omega) = \pi/\omega$.

(d) $\cos\omega t + \cos 3\omega t + \cos 5\omega t$

Step 1: Identify frequencies.

The three terms have angular frequencies $\omega$, $3\omega$, $5\omega$. All are odd multiples of $\omega$.

Step 2: Classify.

This is a sum of cosines with different frequencies. It is periodic (all frequencies are integer multiples of $\omega$, so the sum repeats when $\omega t$ increases by $2\pi$). But it is not a single sine or cosine — it is a Fourier series with three harmonics. So it is periodic but not simple harmonic.

Step 3: Period.

The fundamental frequency is $\omega$, so $T = 2\pi/\omega$. …

Step 1: Classify each function as SHM (single sinusoid $A\sin/\cos(\Omega t+\phi)$), periodic-but-not-SHM (a sum of commensurate sinusoids or a distorted repeating shape), or non-periodic (never repeats), then read off the period from the coefficient of $t$ where relevant.

Step 2 (a): $\sin\omega t-\cos\omega t=\sqrt2\sin(\omega t-\pi/4)$ — single sinusoid, SHM, $T=2\pi/\omega$.

Step 3 (b): The triple-angle identity gives $\sin^3\omega t=\dfrac34\sin\omega t-\dfrac14\sin3\omega t$ — two harmonics ($\omega,3\omega$), so periodic but not SHM; the fundamental period (LCM of $2\pi/\omega$ and $2\pi/3\omega$) is $T=2\pi/\omega$.

Step 4 (c): $3\cos(\pi/4-2\omega t)=3\cos(2\omega t-\pi/4)$ using $\cos(-\theta)=\cos\theta$ — single sinusoid of angular frequency $2\omega$, SHM, $T=\pi/\omega$. …